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What is the remainder when X is divided by 40?

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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 09:14
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IMO A.

Statement 1: 3X + 30 leaves remainder 93 when divided by 120.
The numbers in this form would be 93,213,333,453...
X values for these would respectively would be - 30,70,110,150...
When these values are divided by 40, it always leaves a remainder as 30.
Hence, this statement is sufficient.

Statement 2: 5X - 10 leaves remainder 15 when divided by 20.
The numbers in this form would be 15,35,55,75...
X values for these would respectively would be - 5,9,13,17...
When these values are divided by 40, does not give a fixed value.
Hence, this statement is not sufficient.
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 09:16
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The question states that what is the remainder when X is divided by 40

Statement 1: 3X + 30 leaves remainder of 93 when divided by 120

Let X = 21, therefore, 3(21) + 30 = 93
Thus, 93/120 gives remainder of 93
therefore, 93/40 gives remainder of 13

Let X = 61, therefore, 3(61) + 30 = 213
thus, 213/120 gives remainder of 93
therefore, 213/40 gives remainder of 13

similarly, let X = 101, therefore, 3(101) + 30 = 333
thus, 333/120 gives remainder of 93
therefore, 333/40 gives remainder of 13

Therefore this statement is sufficient (AD)

Statement 2: (5X - 10)/20 gives remainder of 15
let X = 5, therefore (5x5 - 10)/20 = 15/20 gives remainder of 15
therefore, 5/40 gives remainder of 5

let X = 9, therefore (5x9 - 10)/20 = 35/20 gives remainder of 15
therefore, 9/40 gives remainder of 9

Not sufficient

Hence answer choice A.
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 09:19
1
from statement (1),
\(3x + 30 = 120a + 93\)
\(x = 40a + 21\) , so the possible values of \(x\) are (\(21,61,101,...\)),
which all gives \(21\) as a reminder upon dividing by \(40\) --> sufficient

from statement (2),
\(5x - 10 = 20b + 15\)
\(x = 4b + 5\), so the possible values of x are (\(5,9,13,...\))
which gives different values (\(5,9,13,...\)) upon dividing by \(40\) --> insufficient

A
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 09:27
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IMO A

Have to find: Remainder of \(X/40\)

Statement 1 -> \(\frac{(3X + 30)}{120}\) gives a remainder of 93. Substitute for X based on this statement. For X = 21, \(\frac{(3X + 30)}{120}\) gives the remainder of 93 and \(X/40\) gives a remainder of 21. For X = 101, \(\frac{(3X + 30)}{120}\) gives the remainder of 93 and \(X/40\) gives a remainder of 21 again. ---> Sufficient

Statement 2 -> \(\frac{(5X - 10)}{20}\) gives a remainder of 15. Substitute for X based on this statement. For X = 5, \(\frac{(5X - 10)}{20}\) gives the remainder 15 and \(X/40\) gives the remainder 5. For X = 9, \(\frac{(5X - 10)}{20}\) gives the remainder 15, but \(X/40\) gives the remainder 9. ---> Insufficient
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 09:32
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3x +30 will always yield x as 39,79,119 and Soo on

So.remainder will be always 39

But 5x-10 will have values of x different as 13,17 so we have different remainder values

A it is

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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 09:34
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(1) 3X + 30 leaves remainder 93 when divided by 120.

We can express this fraction as

\(\frac{3(x+10)}{120} = n + \frac{93}{120}\) where n is the quotient

This reduces to

\(\frac{(x+10)}{40} = n + \frac{31}{40}\)

So when \(x+10\) is divided by \(40\), the remainder is \(31\)

When \(10\) is divided by \(40\), the remainder is \(10\) so when \(x\) is divided by \(40\), the remainder must be \(21\)

1 is sufficient

(2) 5X - 10 leaves remainder 15 when divided by 20.

Lets express this fraction as \(\frac{5(x-2)}{20} = \frac{m+15}{20}\)

This reduces to \(\frac{(x-2)}{4} = m+\frac{3}{4}\)

x-2 leaves a remainder of 3 when divided by 4

Not sufficient to find out the remainder when x is divided by 40

2 is insufficient

Answer is (A)
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 09:37
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What is the remainder when positive integer X is divided by 40?


STATEMENT (1) 3X + 30 leaves remainder 93 when divided by 120
from this statement, 3X+30 can be written as
3X+30 = 120K + 93-----(K=0,1,2,3,4........)
3(X+10) = 3(40K+31)
X+10 = 40K+31
X=40K+21

X can be 21,61,101,141.........
X divided by 40 gives the remainder 21
so SUFFICIENT

STATEMENT (2) 5X - 10 leaves remainder 15 when divided by 20.
from this statement, 5X-10 can be written as
5X-10 = 20K+15----(K=0,1,2,3,4......)
5(X-2) = 5(4K+3)
X-2 = 4K+3
X = 4K+5

X can be 5,9,13............45,49,53.....
X divided by 40 gives remainder 5,9,13,...
so INSUFFICIENT

A is the answer
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 09:51
1
We can express x=40q+r
Where q is a quotient and r is the remainder (both non-negative integers)
Always note that remainder cant be more than the divisor (in our case 40)

1 stm --> We can insert above value of x to get:
3(40q+r)+30=120q+3r+30
We can observe that 120q is divisible by 120, so 3r+30 divided by 120 will leave 93 as a remainder
We can manipulate with some noble numbers to get 93 remainder when divided by 120 and those lowest integers are 93 and 213
1) 3x+30=93 r=21 Valid as r<40
2) 3r+ 30=213 r=61 Not Valid r>40
and all other integers will give the value of r more than 40.
Only one option is valid r=21

Sufficient

2 stm --> The same process as above
5(40q+r)-10=200q+5r-10 (200q is divisible by 20)
5r-10 divided by 20 leaves remainder 15
1) 5r-10=15 r=5 (r<40)
2) 5r-10=35 r=9 (r<40)

at least two values are possible

Not sufficient

IMO
ANS: A
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 10:01
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We are required to find remainder when \(x\) an unknown positive integer is divided by 40, i.e. the value of "r" where \(x=40m+r\):

(1) 3X + 30 leaves remainder 93 when divided by 120. - This means that \(3x+30=120n+93\) ==>> (divide equation by 3) \(x+10=40n+31\) ==>> \(x=40n+21\). So the remainder of \(\frac{x}{40}\) is 21. Sufficient.

(2) 5X - 10 leaves remainder 15 when divided by 20. - This implies that \(5x-10=20k+15\) ==>> (divide equation by 5) \(x-2=4k+3\)==>>\(x=4k+5\) ==> \(x=4p+1\), meaning that when x is divided by 4, the quotient is an integer p and the remainder is 1. However, we are unable to derive the remainder when \(x\) is divided by 40, insufficient.

Correct answer is A.
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 10:09
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IMO A

What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120.

Say q is quotient, then from st.1 => 3X + 30 = 120q + 93 => 3X = 120q + 63 => X = 40q + 21
Clearly, remainder is 21 when X is divided by 40
Sufficient

(2) 5X - 10 leaves remainder 15 when divided by 20.

Say q is quotient, then from st.2 => 5X - 10 = 20q + 15 => 5X = 20q + 25 => X = 4q + 5
Now from this equation we can get X = 5, 9, 13, 17, 21, etc.
And if 5 is divided by 40, the remainder is 5, but if 9 is divided by 40, the remainder is 9
So, we have different values.
Insufficient
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 10:12
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(1) 3X + 30 leaves remainder 93 when divided by 120.
X will be 21, 61 and so on, remainder will always be 21. Sufficient

(2) 5X - 10 leaves remainder 15 when divided by 20.
Different remainders for the values of X. Not Sufficient.
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 10:20
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(1) 3X + 30 leaves remainder 93 when divided by 120.
That means 3x+30-93 is divisible by 120
3x-63 = 120a (where 'a' is a positive integer)
3(x-21)=120a
x-21 = 40a
so we can determine the remainder as 21 when divided by 40. Sufficient

(2) 5X - 10 leaves remainder 15 when divided by 20.
5x-25 is divisible by 20
5(x-5)=20a
x-5=4a
We will not be able to calculate remainder when x is divisible by 40 from this equation. Insufficient.

IMO A
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 10:20
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Quote:
What is the remainder when positive integer X is divided by 40?


X is integer and x > 0
We need to identify if it is possible to find a remainder when X is divided by 40 in this Data Sufficiency (DS) question:
Let us analyze the statements:

Statement 1:
(1) 3X + 30 leaves remainder 93 when divided by 120.

Let us write it down as a formula: \(3X + 30 = 120*a + 93\), where a is the number of times 120 is repeated in 3X + 30
\(3X = 120*a + 93 - 30 = 120*a + 63 = 3 * (40*a + 21)\)
\(X = 40*a + 21\)
Thus, 21 is remainder when X is divided by 40.

Sufficient.

Statement 2:
(2) 5X - 10 leaves remainder 15 when divided by 20.

Again, let us represent it with the formula: \(5X - 10 = 20*b + 15\), where b is the number of times 20 is repeated in 5X - 10
\(5X = 20*b + 15 + 10 = 20*b + 25 = 5 * (4*b + 5)\)
\(X = 4*b + 5\)
Unfortunately, we do not possess information whether b is equal to 1, 10 or 15, and for this reason are unable to answer the question about the remainder using this statement alone.

Insufficient.

Answer: A
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 10:39
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What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120. --> correct: 3X+30 = 120*d + 93 => X+10 = 40*d + 31 => X = 40 *d + 21, so if X is divided by 40, the reminder will be 21
(2) 5X - 10 leaves remainder 15 when divided by 20. --> 5X -10 = 20 * D + 15 => X -2 = 4 * D + 3 => X = 4 * D + 5 => X = 4* (D+1) + 1, now case-1: if X =61, then if 61 divided by 4, reminder will be 1 & if 61 divided by 40, reminder will be 21 but case-2: if X =41, then if 41 divided by 4, reminder will be 1 & if 41 divided by 40, reminder will be 1. So different reminder for different cases.

SO the Answer is A
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 10:44
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To find,

Remainder of X when X is divided by 40.

We know,

Dividend = Divisor * Quotient + Remainder

Let us check the two options -

Option 1: 3X + 30 leaves remainder 93 when divided by 120.

=> 3*X + 30 = 120 * Quotient + 93
=> X + 10 = 40 * Quotient + 31
=> X = 40 * Quotient + 21

Hence, X whenever divided by 40 will give a remainder of 21.

Hence option 1 is sufficient.

Option 2: 5X - 10 leaves remainder 15 when divided by 20.

=> 5*X - 10 = 20 * Quotient + 15
=> X - 2 = 4 * Quotient + 3
=> X = 4 * Quotient + 5
=> X = 4 * Quotient + 4 + 1
=> X = 4 * (Quotient + 1) + 1

Hence, X whenever divided by 4 will give a remainder of 1.

Now, if X is 5 (satisfying above condition in option 2) then it will give a remainder of 5 when divided by 40.
Now, if X is 9 (satisfying above condition in option 2) then it will give a remainder of 9 when divided by 40.
Now, if X is 45 (satisfying above condition in option 2) then it will give a remainder of 5 when divided by 40.
Now, if X is 53 (satisfying above condition in option 2) then it will give a remainder of 13 when divided by 40.

Hence, option 2 is insufficient.

Answer: A
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 11:02
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IMO-A

Remainder when positive integer X is divided by 40

(1) 3X + 30 leaves remainder 93 when divided by 120.

3X+30= 120K + 93
=> X+10= 40K + 31 [K = integer]
=> X-21 = 40K
=> X-21= { 0, 40, 80, ........40K}
=> X= 21, 61, 81,....40K+21
Remainder (X/40)= 21

Sufficient

(2) 5X - 10 leaves remainder 15 when divided by 20.
=>.5X-10=20K+15
=> 5X= 20K+25
=> X= 4K+5
=> X= {5,9,13,17...........4K+5}
Remainder (X/40)= {5,9,13,17.....}

Not Sufficient
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 11:55
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What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120.
(2) 5X - 10 leaves remainder 15 when divided by 20.

Stmt 1: if x =61, then 3*61+30 = 213 , which results in 93 when divided by 120
if x =141, then 3*141+30 = 453 , which results in 93 when divided by 120.
Different values of X give similar remainder. so sufficient.

Stmt 2: if x =9, then 5*9-10 = 35 , which results in 15 when divided by 20
if x =13, then 5*13-10 = 55 , which results in 15 when divided by 20,
Different values of X give different remainder. so insufficient.

So, the correct answer choice is (A)
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 12:19
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What is the remainder when positive integer X is divided by 40?

40q+r = X ..... q is quotient and r is remainder. We have to find r.

(1) 3X + 30 leaves remainder 93 when divided by 120.

120q+93 = 3X + 30
120q+63 = 3X
40q+21 = X ... This is in the same form as 40q+r=X

Therefore r = 21

(1) IS SUFFICIENT

(2) 5X - 10 leaves remainder 15 when divided by 20.

20q+15 = 5X-10

20q + 25 = 5X

4q + 5 = X

It is not possible to represent X in the form of 40q+r....

X could be 9 with remainder 9, or x could be 45 with remainder 5.

(2) IS NOT SUFFICIENT

ANSWER: A - 1 Alone is SUFFICIENT
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What is the remainder when X is divided by 40?  [#permalink]

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New post Updated on: 18 Jul 2019, 04:43
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What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120.
(2) 5X - 10 leaves remainder 15 when divided by 20.

condition 1,

3x + 30 leaves remainder 93, when divided by 120

assume first no 93, so 3x is 63 hence x is 21. so remainder is 21 if divided by 40

next x will be increased by 120/ coefficient of x ie 120/3 is 40. because other no is just an addition which will always maintain the same distance
so next x is 61 which gives , 183 + 30 is 213, if divided by 120 will leave remainder 93 and so on..
so x can be 21,61,101.... so on . so whatever the remainder is 21 when divided by 40
so this is sufficient, ans is always 21.

condition 2,

5x-10 , remainder 15. when divided by 20


assume n = 15, which gives, 5x = 25 so x =5. , so remainder if divided by 40 is 5

next x will be increased by 20/ coefficient of x ie 20/5 which is 4.
so next x is 5 + 4 is 9, which gives, 45-10, 35 if divided by 20 will leave 15 and so on

x can be 5,9,13,17 .... so on and each leaves different remainder when divided by 40 so clearly insufficient

so ans is A
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Originally posted by ccheryn on 17 Jul 2019, 12:50.
Last edited by ccheryn on 18 Jul 2019, 04:43, edited 3 times in total.
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Re: What is the remainder when X is divided by 40?  [#permalink]

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New post 17 Jul 2019, 12:51
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What is the remainder when positive integer X is divided by 40?

(1) 3X + 30 leaves remainder 93 when divided by 120.
Statement 1 can be written as 3X+30 = 120k+93
Dividing by 3 we get X+10=12k+31 --> X=40k+21.
Dividing by 40 we get X/40 = k+21/40. Remainder is 21.
Hence Statement 1 is sufficient.

(2) 5X - 10 leaves remainder 15 when divided by 20.
Statement 2 can be written as 5X-10=20k+15 --> 5X=20k+25
Dividing by 5 we get X=4k+5
Dividing by 40 we get X/40 = (4k+5)/40
If k = 0, we get remainder as 5.
If k = 1, we get remainder as 9.
Therefore we don't get fixed remainder in this case.
Hence statement 2 is not sufficient.


Statement 1 alone is sufficient.

Answer Choice: A
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Re: What is the remainder when X is divided by 40?   [#permalink] 17 Jul 2019, 12:51

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